Fine the largest domain for f(x) = (x-1)^2, with upper x bound 10, such that the function is one to one and non decreasing

Respuesta :

Answer:

*look at the step by step*

Step-by-step explanation:

If you plot this on a graphing calculator or Desmos you will see it is always 1-1 (for a given x, one value of y)

The function is zero when x = 7, so touches the x axis at (7,0).  To the left of (7,0) funcon is decreasing (as x increases, y decreases), to the right of (7,0) the function is increasing.  

so the domain (x values) where f is increasing is  x >7  or  (7, +∞)

Range of f (possible y values)  is [0,+∞)

for the inverse

f (x) = (x-7)2

lets put f(x) = y

y = (x-7) 2  

to find the inverse function

get x in terms of y

switch the x and y

y = (x-7)2

√y  =  x - 7

7 + √y     =  x

switch x, y

7 + √x  = y

y = f-1 (x)

f-1  (x) =  7 + √x

Domain of the inverse :    f-1 (x) will exist as long as x >= 0,  (so the square root exists) so the domain should be [0, + ∞).   However the question states the inverse is restricted to the domain above, so domain is x > 7  or  (7, +∞).

Range of  the inverse.   Obviously is is strictly increasing (as x increases , y increases) so the minimum values must be 7 + √7, and maximum + ∞.          So range is  ( 7 + √7, + ∞)

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